We do not actually know the population proportion unless we conduct an expensive poll of all individuals in the population. Our earlier value of p = 0.88 was based on a Pew Research conducted a poll of 1000 American adults that found p̂ = 0.887 of them favored expanding solar energy. The researchers might have wondered: does the sample proportion from the poll approximately follow a normal distribution? We can check the conditions from the Central Limit Theorem:
Independence. The poll is a simple random sample of American adults, which means that the observations are independent.
Success-failure condition. To check this condition, we need the population proportion, p, to check if both np and n(1− p) are greater than 10. However, we do not actually know p, which is exactly why the pollsters would take a sample! In cases like these, we often use p̂ as our next best way to check the success-failure condition:
np̂ = 1000× 0.887 = 887 n(1− p̂) = 1000× (1− 0.887) = 113
The sample proportion p̂ acts as a reasonable substitute for p during this check, and each value in this case is well above the minimum of 10.
This substitution approximation of using p̂ in place of p is also useful when computing the standard error of the sample proportion:
√ p(1− p)
n ≈ √ p̂(1− p̂)
√ 0.887(1− 0.887)
1000 = 0.010
This substitution technique is sometimes referred to as the “plug-in principle”. In this case, SEp̂ didn’t change enough to be detected using only 3 decimal places versus when we completed the calculation with 0.88 earlier. The computed standard error tends to be reasonably stable even when observing slightly different proportions in one sample or another.
5Since the sample size n is in the denominator (on the bottom) of the fraction, a bigger sample size means the entire expression when calculated will tend to be smaller. That is, a larger sample size would correspond to a smaller standard error.
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